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julia-pushkina [17]
3 years ago
15

Are fossils likely to form in areas that experience a lot of erosion on daily basis?

Chemistry
1 answer:
frosja888 [35]3 years ago
6 0
No, There are not likely to form in areas that experience a lot of erosion on a daily basis.
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What is the treatment and disposal of Mercury<br>​
PIT_PIT [208]
<h2>Hey there! :) </h2>

<h3>The treatment and disposal of Mercury:</h3>

  • Heating and incineration can release the mercury vapor into atmosphere causing atmospheric pollution. The process of solidification and disposal into secured landfill, gas phase recovery of mercury, and thermal treatment is gaining interest in mercury treatment and recovery field by various researchers and industries.
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4 0
3 years ago
Which statement is true for most autotrophic organisms?
lara [203]

omg yees

eExplanation:

5 0
3 years ago
How many sucrose molecules will be in 250 grams
sasho [114]
<h2>Answer:</h2>

We will need to know Avogadro's number and the molar mass of sucrose for this problem to do dimensional analysis.

  • Avogadro's number: 6.022 × 10²³ molecules
  • Molar mass of sucrose: 342.2965 g/mol

250g × \frac{1 mol}{342.297g} × \frac{6.022 * 10^{23} molecules}{1 mol} = 4.398 molecules

There are <em>4.398 sucrose molecules </em>in 250 grams of sucrose.

6 0
3 years ago
Which statement best describes the work and energy in these examples?
kari74 [83]

Answer:

The gas and food are examples of energy.

Explanation:

The reason why is that food is energy as proteins and electrolytes. Gas is a energy of fossil fuels.

3 0
3 years ago
This reaction was at equilibrium when 0.2 atm of iodine gas was pumped into the container, what happened to the equilibrium and
n200080 [17]

Answer:

Q was < K. Partial pressure of hydrogen decreased, iodine increased

Explanation:

<em>After iodine was added the Q was [Select] K so the reaction shifted toward the Products [Select] ,The partial pressure of hydrogen [Select], Iodine [Select] |,and hydrogen iodide Decreased</em>

Based on the equilibrium:

H2(g) + I2(g) ⇄ 2HI(g)

K of equilibrium is:

K = [HI]² / [H2] [I2]

<em>Where [] are concentrations at equilibrium</em>

And Q is:

Q = [HI]² / [H2] [I2]

<em>Where [] are actual concentrations of the reactants.</em>

<em />

When the reaction is in equilibrium, K=Q.

But as [I2] is increased, Q decreases and Q was < K

The only concentration that increases is [I2], doing partial pressure of hydrogen decreased, iodine increased

7 0
3 years ago
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